# find square roots of -1+2i We have to find the square root of -1+2i i.e. \sqrt{-1+2i}

We will find the square roots of the complex number of the form x+yi , where x and y are real numbers, by the following method:

Let z^2=(x+yi)^2=-1+2i

i.e. (x^2-y^2)+2xyi=-1+2i

Comparing real and imaginary terms we get,

x^2-y^2=-1 ------->...

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We have to find the square root of -1+2i i.e. \sqrt{-1+2i}

We will find the square roots of the complex number of the form x+yi , where x and y are real numbers, by the following method:

Let z^2=(x+yi)^2=-1+2i

i.e. (x^2-y^2)+2xyi=-1+2i

Comparing real and imaginary terms we get,

x^2-y^2=-1 -------> (1)

2xy=2 implies xy=1 ------>(2)

So from (2) we get,  y=1/x . Substituting this in (1) we have,

x^2-\frac{1}{x^2}=-1

i.e. x^4+x^2-1=0

implies x^2=\frac{-1\pm\sqrt{5}}{2}

=0.62, -1.62

Therefore, x=\pm\sqrt{0.62}=\pm 0.79 

x^2=-1.62 is discarded since it gives imaginary value.

hence,

When x=0.79,  y= 1.27

x=-0.79 , y= -1.27

i.e we have,  \sqrt{-1+2i}=0.79+1.27i or -0.79-1.27i

=\pm (0.79+1.27i)

Hence the square roots of -1+2i are:  \pm` (0.79+1.27i)

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